Problem: Express $z_1=3\sqrt{3}-9i$ in polar form. Express your answer in exact terms, using radians, where your angle is between $0$ and $2\pi$ radians, inclusive. $z_1=$
Answer: The Strategy A complex number in rectangular form, $z={a}+{b}i$, can be written in polar form as $z={r}[\cos{\theta}+i\sin{\theta}]$, where ${r}$ is the absolute value, or modulus, and ${\theta}$ is the angle, or argument. Therefore, ${r}$ and ${\theta}$ can be found using the following formulas: ${r}=\sqrt{{a}^2+{b}^2}$ $\tan{\theta}=\dfrac{{b}}{{a}}$ [How did we get these equations?] Similarly, a complex number in polar form, $z={r}[\cos{\theta}+i\sin{\theta}]$, can be written in rectangular form as $z={a}+{b}i$, using the following formulas: ${a}={r}\cos{\theta}$ ${b}={r}\sin{\theta}$ [How did we get these equations?] Finding $r$ For $z_1={3\sqrt{3}}{-9}i$ : ${a} = {3\sqrt{3}}$ ${b} = {-9}$ Therefore, we can find ${r}$ as follows. $\begin{aligned}{r}&=\sqrt{{a}^2+{b}^2} \\\\&=\sqrt{({3\sqrt{3}})^2+({-9})^2} \\\\&=\sqrt{27+81} \\\\&={\sqrt{108}} \\\\&={6\sqrt{3}}\end{aligned}$ Finding $\theta$ Using the formula, we have: $\begin{aligned}{\theta}&=\arctan\left(\dfrac{{b}}{{a}}\right) \\\\&=\arctan\left(\dfrac{{-9}}{{3\sqrt{3}}}\right) \\\\&={-\dfrac{\pi}{3}}\end{aligned}$ Since ${a}$ is positive and ${b}$ is negative, ${\theta}$ must lie in Quadrant $\text{IV}$. Therefore its angle must be between $\dfrac{3\pi}{2}$ and $2\pi$ radians. Using the identity $\tan(2\pi+\theta)=\tan(\theta)$, we know that the following is also a solution of the equation. $2\pi+(-\dfrac{\pi}{3})=\dfrac{5\pi}{3}$ So $\theta = {\dfrac{5\pi}{3}}$. Summary $z_1={6\sqrt{3}}\left[\cos{\dfrac{5\pi}{3}}+i\sin{\dfrac{5\pi}{3}}\right]$